A) \[4{{a}^{2}}{{b}^{2}}\]
B) \[{{a}^{2}}-{{b}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[-{{a}^{2}}{{b}^{2}}\]
Correct Answer: C
Solution :
We have \[\sin (\alpha -\beta )=\sin (\theta -\beta -\overline{\theta -\alpha })\] \[=\sin (\theta -\beta )\cos (\theta -\alpha )-\cos (\theta -\beta )\sin (\theta -\alpha )\] \[=ba-\sqrt{1-{{b}^{2}}}\sqrt{1-{{a}^{2}}}\] and \[\cos (\alpha -\beta )=\cos (\theta -\beta -\overline{\theta -\alpha })\] \[=\cos (\theta -\beta )\cos (\theta -\alpha )+\sin (\theta -\beta )\sin (\theta -\alpha )\] \[=a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}}\] \[\therefore \] Given expression is \[{{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta )\] \[=(a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}{{)}^{2}}}+2ab\{ab-\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}\}\] \[={{a}^{2}}+{{b}^{2}}\]. Trick : Put \[\alpha =30{}^\circ ,\beta =60{}^\circ \] and \[\theta =90{}^\circ ,\] then \[a=\frac{1}{2},b=\frac{1}{2}\] \[\therefore {{\cos }^{2}}(\alpha -\beta )+2ab\sin (\alpha -\beta )=\frac{3}{4}+\frac{1}{2}\times \left( -\frac{1}{2} \right)=\frac{1}{2}\] which is given by option (c).
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